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This is a problem in my exam and I can't find the solution using elementary inequality knowledge. Can anyone here help me solve this. Thanks

$a,b,c $ are positive real numbers which satisfy $(a+c)(b+c) = 4c^{2}$. Find the minimum of this expression: $$P = \frac{32a^{3}}{(b+3c)^{3}} + \frac{32b^3}{(a+3c)^{3}} - \frac{\sqrt{a^{2} + b^{2}}}{c}$$

Thanks so much.

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We begin with the observation that the change $a=xc, b=yc$ reduces the problem under consideration to the two-dimensional optimization problem $$\min \frac {x^3} {(y+3)^3} + \frac {y^3} {(x+3)^3}-\sqrt{x^2+y^2}$$ under the constraints $$(x+1)(y+1)=4, x \ge 0, y \ge 0 . $$ Solving it with Mathematica by

Minimize[ 32 x^3/(y + 3)^3 + 32 y^3/(x + 3)^3 - Sqrt[x^2 + y^2], 
 (x + 1)*(y + 1) == 4 && x >= 0 && y >= 0, {x, y}]

we obtain:

{1 - \sqrt{2}, {x -> 1, y -> 1}}

i.e. $\{a=b,b=c\}.$ One may play with Lagrange multipliers to this end.


Addition. For each nonnegative $x$ and for each nonnegative $y$ the inequality $$ \frac {x^3} {(y+3)^3} + \frac {y^3} {(x+3)^3}-\sqrt{x^2+y^2} \ge 2\sqrt{\frac {x^3y^3} {(x+3)^3(y+3)^3}}-\sqrt{x^2+y^2}$$ holds. The equality takes place iff $\frac {x^3} {(y+3)^3} =\frac {y^3} {(x+3)^3}.$ This implies $x=y$ because the function $f(x):=x^3(x+3)^3$ increases for nonnegative $x$. Taking into account $(x + 1)(y + 1) =4$, we obtain the optimal solution $\{x=1, y=1\}$.

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  • $\begingroup$ Is there any "pure" solution for this, such as using primary popular inequality? Your solution is clearly right, but it's not proper for high school student... $\endgroup$ – le duc quang Jul 4 '13 at 15:21
  • $\begingroup$ If a high school student knows elements of calculus, then the one can substitute $y= \frac 4 {x+1} -1$ and find the minimum of a function of single variable $x$. An elementary solution was not demanded in your question. $\endgroup$ – user64494 Jul 4 '13 at 16:04
  • $\begingroup$ OH sorry, it's my fault. I editted my question. Can you try to think about it? I really appreciate. $\endgroup$ – le duc quang Jul 4 '13 at 16:09
  • $\begingroup$ Can you specify what you mean by elementary popular inequalities? $\endgroup$ – user64494 Jul 4 '13 at 16:22
  • $\begingroup$ I mean we can use popular inequality such as Cauchy, B-C-S,...I don't understand what you mean by "Solving it with Mathematica". Is this a tool or an algorithm.One more thing, the degree of denominator in my problem is 3, not 1 as you rewrite in line 3. $\endgroup$ – le duc quang Jul 4 '13 at 17:44

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